How do you integrate $(x^2e^(x/2))dx$ ?

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Answer 1 · 2017-01-23T20:11:30

$intx^2e^(x/2)dx=2e^(x/2)(x^2-4x+8)+C$

$I=intx^2e^(x/2)dx$

Use the substitution $t=x/2$ . This implies that:

${(2t=x),(2dt=dx),(4t^2=x^2):}$

Then:

$I=intx^2e^(x/2)dx=int(4t^2)(e^t)(2dt)=int8t^2e^tdt$

Now use integration by parts. This takes the form $intudv=uv-intvdu$ . For the current integral, let:

${(u=8t^2" "=>" "du=16tdt),(dv=e^tdt" "=>" "v=e^t):}$

So:

$I=uv-intvdu=8t^2e^t-int16te^tdt$

Again perform integration by parts with a knew $u$ and $dv$ :

${(u=16t" "=>" "du=16dt),(dv=e^tdt" "=>" "v=e^t):}$

Hence:

$I=8t^2e^t-(uv-intvdu)$

$I=8t^2e^t-16te^t+int16e^tdt$

Note that $int16e^tdt=16inte^tdt=16e^t$ :

$I=8t^2e^t-16te^t+16e^t+C$

$I=8e^t(t^2-2t+2)+C$

Returning to $x$ from the substitution $t=x/2$ :

$I=8e^(x/2)(x^2/4-x+2)+C$

$I=2e^(x/2)(x^2-4x+8)+C$

How do you integrate (x^2e^(x/2))dx(x^2e^(x/2))dx?